Question

The density of ${\text{CaO}}$ is 3.35 ${\text{gm/c}}{{\text{m}}^{\text{3}}}$. The oxide crystallizes in one of the cubic systems with an edge of 4.80 Å. how many $C{a^{2 + }}$ ions belong to each unit cell?[${\text{CaO}}$= 56]A.3B.4C.6D.2

Hint: Density of a unit cell can determine the value of formula units as it is the ratio of mass and volume of unit cell. Mass of the unit cell is equal to the mass of each atom in the unit cell and the number of atoms in the unit cell.
Formula used:
$\rho {\text{ }} = {\text{ }}\dfrac{{ZM}}{{{N_A}{{(a)}^3}}}$

As the density is given in the question, we can use the below formula to find the number of ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ions that belong to each unit cell.
$\rho {\text{ }} = {\text{ }}\dfrac{{ZM}}{{{N_A}{{(a)}^3}}}$
To calculate Z, this can be rearranged as $Z{\text{ }} = {\text{ }}\dfrac{{\rho {N_A}{{(a)}^3}}}{M}$
Given: molar mass of, M= 56 g/mole
Density, ${{\rho }}$: 3.35 $gm/c{m^3}$
Edge length, $\; = {\text{ }}4.80$$\times {\text{ }}{10^{ - 8}}$cm
Avogadro number,${{\text{N}}_{\text{A}}}$$= {\text{ }}6.022$${\text{ \times 1}}{{\text{0}}^{{\text{23}}}}$
Putting all these values in the below equation, we get
$Z{\text{ }} = {\text{ }}\dfrac{{\rho {N_A}{{(a)}^3}}}{M}$
$Z{\text{ }} = {\text{ }}\dfrac{{(3.35{\text{ }}gm/c{m^3})(6.022 \times {\text{ }}{{10}^{23}}){{(4.80 \times {\text{ }}{{10}^{ - 8}}cm)}^3}}}{{56{\text{ }}g/mole}}$
$\therefore {\text{ }}Z{\text{ }} = {\text{ }}3.98{\text{ }} \approx {\text{ }}4$formula units

Hence, the correct option is (B).

Note:
The cubic system that has ${\text{Z = }}$4 belongs to a face centred cubic lattice. Therefore, ${\text{CaO}}$belongs to the FCC lattice.
The value for the formula unit determines the type of crystal structure the given molecule belongs to. For BCC, ${\text{Z = }}$2 and for simple cubic or primitive cells, ${\text{Z = }}$1.